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Assume the zero set $A=\{s\in [0,t]:g(s)=0\}$ is measurable, as is ensured by measurability of $g$. This is an implicit necessary hypothesis: for an arbitrary real function the printed integral and event need not be measurable. Put $a={\int }_0^t{1}_A(s)ds$. Conditional on $N_t=n$, the uniform order statistics description makes the probability of no jump in $A$ equal to $(1-a/t{)}^n$. Averaging over the Poisson distribution givesConsequentlyIn particular a measure-zero zero set is almost surely missed, despite possibly containing infinitely many points.

Use the slotted ALOHA model with one outstanding packet per active station and a collision channel: a slot with exactly one attempt succeeds, whereas an idle slot or a slot with two or more attempts delivers nothing. Let $Y_t$ be independent Poisson random variables of mean $\nu$, independent also of the transmission choices. Arrivals join after the current slot's attempts. Given $(S_t,N_t)=(S,N)$, the attempt count has a binomial distribution with parameters $N$ and $1/S$. Consequently, for $N\ge 1$,For $N=0$, set $p_0=1$ and $p_1=0$. The equation $N_{t+1}=N_t+Y_t-1_{\{Z_t=1\}}$ adds arrivals and removes precisely one packet following a successful slot. The estimate $S_t\ge 1$ controls the common transmission probability and updates using the idle, success, or collision feedback. The conditional distribution of the next state depends only on $(S_t,N_t)$, so this pair is a Markov chain.

Away from the reflecting floor in $S$, its one-step conditional expectations areWhen $N$ and $S$ are large with $N/S\approx \kappa$, the Poisson limit theorem gives $p_0\approx e^{-\kappa }$ and $p_1\approx \kappa e^{-\kappa }$. Treating these approximate drifts as derivatives motivates the fluid approximation of slotted ALOHA:The floor at $S=1$ is negligible in this interior scaling. This calculation motivates the ordinary differential equations; it is not a proof of a stochastic fluid limit.

For $(a,b,c)=(2-e,0,1)$, putThen $g^{\prime }(\kappa )=(e-2+\kappa )e^{-\kappa }>0$ and $g(1)=0$. On the domain $s>0$, introduce time $\tau$ by $d\tau /dt=1/s$. Differentiating $\kappa =n/s$ giveswhere $r(\kappa )=\kappa [1-(e-2+\kappa )e^{-\kappa }]$.

The function $r$ is strictly increasing from zero to infinity, and $r(1)=e^{-1}$. To check the monotonicity explicitly,Indeed, with $d=3-e\in (0,1)$,because its discriminant is $d^2-4d<0$; and $e^{\kappa }\ge 1+\kappa$. Thus the scalar ordinary differential equation for $\kappa$ points toward the unique value ${\kappa }_{\ast }$ satisfying $r({\kappa }_{\ast })=\nu$. Every trajectory with $s(0)>0$, $n(0)\ge 0$ has bounded $\kappa$ and $\kappa (\tau )\to {\kappa }_{\ast }$. Positivity of $s$ follows from $s(\tau )=s(0)\exp ({\int }_0^{\tau }g(\kappa (u))du)$.

If $0\le \nu <e^{-1}$, then ${\kappa }_{\ast }<1$ and $g({\kappa }_{\ast })<0$. Thus $s(\tau )$ decays exponentially for large $\tau$, andThis is finite-time draining of a fluid model. The unmodified equations have $n/s$ undefined at the origin, so the statement that trajectories converge to the origin needs this endpoint interpretation; continuing with an absorbing zero trajectory is an additional fluid-model convention, not a solution of the displayed equations at the origin.

If $\nu >e^{-1}$, then ${\kappa }_{\ast }>1$ and $g({\kappa }_{\ast })>0$. Now $s(\tau )$ grows exponentially, physical time tends to infinity, andIn particular the trajectory escapes to infinity. Even without the ratio analysis, $\dot{n}\ge \nu -e^{-1}>0$, since ${\max }_{\kappa \ge 0}\kappa e^{-\kappa }=e^{-1}$.

For the actual Markov chain, it is essential not to replace the finite-$N$ success probability by $e^{-1}$: a single backlogged packet can succeed with probability one. Instead, for $N\ge 2$, maximizing over $S\ge 1$ gives the ALOHA throughput boundIf $\nu >e^{-1}$, choose $K_0$ and $r_0<\nu$ so that $p_1(N,S)\le r_0$ for all $N>K_0$, uniformly in $S$. For sufficiently small $\theta >0$,because its derivative at zero is $r_0-\nu <0$. Independence of arrivals and attempts giveswhenever $N>K_0$. For any integer $K\ge K_0$, stop at the stopping time ${\sigma }_K=\inf \{t:N_t\le K\}$. The stopped process $e^{-\theta N_{t\wedge {\sigma }_K}}$ is a nonnegative supermartingale. The optional stopping theorem, first at bounded times and then by a limit, yields the exponential-supermartingale escape boundFrom every state with $N\le K$, a Poisson distribution arrival burst has a uniformly positive probability of sending the next backlog to at least $K+m$, for any fixed $m>0$. From there the probability of never revisiting $N\le K$ is at least $1-e^{-\theta m}$. By the Markov property at successive visits, the probability of infinitely many visits to this strip is zero: each visit has a uniform positive chance of being the last. This holds for every integer $K\ge K_0$, soTherefore the Markov chain is transient for every finite choice of $a,b,c$. The proof is uniform in $S$ and does not infer stochastic transience merely from the approximate ordinary differential equations.

Let $N$ be the input photoelectron count and $\mu =\eta Q$ its mean, with quantum efficiency $\eta$ and mean incident photon count $Q$. Independent arrivals give a Poisson distribution, so $\mathrm{Var}N=\mathbb{E}N=\mu$. Conditional on $N$, the preceding gamma distribution gives $\mathbb{E}[X\mid N]=gN$ and $\mathrm{Var}(X\mid N)=g^{2}N$. The law of total variance therefore yieldsOne contribution is the ordinary photon shot noise; the other is multiplication excess noise. Thus, neglecting read noise and backgrounds,This is an excess-noise factor $\sqrt{2}$ in analog operation: actual photon conversion efficiency has not been halved. At low occupancy, thresholding each pixel as zero or one event can avoid most multiplication noise, but high arrival rates produce coincident events that cannot be counted separately. That is why the high-rate result concerns charge measurement rather than ideal binary photon counting.

The normalized ansatz intended in the question isfor which $\varphi (1,t)=1$. Substitution into the equation from part (ii) and cancellation of $(s-1)\varphi$ gives the linear ordinary differential equationThusThis generating function is that of a Poisson distribution with parameter $f(t)$. Equivalently, differentiating at $s=1$ givesConsequently

The number probabilities area Poisson distribution of parameter $|z{|}^2$. HenceThe relative uncertainty is $1/|z|=1/\sqrt{⟨N⟩}$, which tends to zero as the mean occupation tends to infinity.

With state-dependent admission, the effective birth rate in state $n$ iswhile the death rate is $\mu$ for $n\ge 1$. The stationary ratios for an M-M-1 queue with state-dependent admission are thereforeInduction gives ${\pi }_n={\pi }_0{\rho }^n/n!$, and normalization identifies a Poisson distribution of mean $\rho$:

Ignoring terms independent of $\theta$, the log-likelihood isIts score function isSince ${\ell }^{\prime \prime }(\theta )=-({\sum }_i{Y}_i)/{\theta }^2\le 0$, the interior critical point is the maximum, givingThe sum of independent Poisson distributions is Poisson with mean $\theta {\sum }_i{x}_i$, soThus the maximum-likelihood estimator is also unbiased.

The probability generating function of each $X_i$ is $\exp (z-1)$. Independence makes the generating function of $S_n$ equal to $\exp (n(z-1))$, so the addition of independent Poisson random variables gives $S_n\sim \mathrm{Pois}(n)$.

The Poisson distribution has mean and variance $n$, hence $\mathbb{E}{Y}_n=0$ and $\mathbb{E}[Y_n^2]=1$. Since $Y_n^{-}\le |Y_n|$, Chebyshev inequality gives

Put $p_i=\mathbb{P}(X_i=1)$ and $\lambda ={\sum }_i{p}_i$. A relative-entropy form of the Poisson approximation bound for dependent Bernoulli variables isThe last two terms form the total correlation; they vanish when the Bernoulli variables are independent.

Here $D_e$ and $H_e$ use natural logarithms.

To prove the bound, let $Q_i$ be the Poisson distribution with mean $p_i$ and let $Q={⨂}_i{Q}_i$. Expanding the Kullback-Leibler divergence against this product law givesThe supplied one-dimensional estimate bounds the first sum by ${\sum }_i{p}_i^2$. Under the addition map, $P_{X_1^n}$ becomes $P_{S_n}$, while the Sum of independent Poisson random variables under $Q$ has the Poisson distribution with mean $\lambda$. The data processing inequality for relative entropy proves the displayed result.

If a bound directly in the paper's unhalved total-variation norm is desired, Pinsker's inequality also gives

Condition on $N_t=n$. By Poisson process conditional arrival times, the jump times are the ordered version of independent $U_1,\dots ,U_n$ uniformly distributed on $[0,t]$. The sum is symmetric in the marked pairs, and the marks are independent, soWrite the bracket as $a$. Averaging over the Poisson distribution of $N_t$ givesThis is the exponential formula for a marked Poisson sum.

In Kendall notation, the first two letters $M$ mean Poisson arrivals and exponential service times. An M-M-1 queue has arrival rate $\lambda$, one server of rate $\mu$, and queue-length transitionsIt has a stationary distribution of an M-M-1 queue exactly whenand then

An M-M-infinity queue has infinitely many servers, so every customer begins service immediately. Its occupancy transitions areThe detailed balance for a birth-death process equations giveThus for every $\lambda ,\mu >0$ the stationary distribution of an M-M-infinity queue exists and isthe Poisson distribution of mean $\rho$. The Burke theorem for an M-M-infinity queue states that in this stationary regime the departure process is a Poisson process of rate $\lambda$; equivalently, time reversal turns departures into arrivals. The past departure process is also independent of the number in the system at the observation time.

For an M-M-1 queue, the first service lasts on average $1/\mu$. During it, the mean number of arrivals is $\lambda /\mu$, and each arrival initiates another statistically identical busy-period family. Thus, when $\lambda <\mu$,so the mean busy period of an M-M-1 queue isFor $\lambda =\mu$ the busy period is almost surely finite but has infinite mean, while for $\lambda >\mu$ it is infinite with positive probability; in either case its expectation is infinite.

For an M-M-infinity queue, stationary occupancy is Poisson with mean $\rho =\lambda /\mu$, so the long-run probability of an empty system isThe process alternates between idle periods and busy periods. An idle period waits for the next Poisson arrival, so its mean is $1/\lambda$. The renewal-reward theorem, with reward equal to time spent empty, givesSolving yields the mean busy period of an M-M-infinity queue:

A one-parameter exponential family with natural statistic $Y$ has density or probability mass functionwhere $\theta$ is the natural parameter of an exponential family and $K$ is the cumulant function of an exponential family. Its mean parameter is

The exponential-family deviance from ${\theta }_1$ to ${\theta }_2$ is twice the Kullback-Leibler divergence:The carrier $h(Y)$ cancels from the likelihood ratio, so, with ${\mu }_1={\mathbb{E}}_{{\theta }_1}Y$,

For the Poisson distribution with mean $\mu$,ThereforeWriting ${\mu }_2={\mu }_1+\delta$ and applying the Taylor series of $\log (1+\delta /{\mu }_1)$ givesso the second-order approximation is

Writing $Y_i$ for team1_goal and $x_i$ for team1_xg, the first hypothesis concerns only the expected value:The second specifies the entire conditional probability distribution:with independence between matches as assumed in the question. The Poisson limit theorem, often called the law of small numbers in this setting, suggests a Poisson distribution when a goal count is the sum of many approximately independent rare scoring opportunities.

Under $H_1$, the slope in this through-origin linear regression is $1$. Its observed Wald statistic isThe corresponding two-sided p-value is about $0.07$, so the fit does not give evidence against $H_1$ at the $5\%$ significance level.

The usual least-squares standard error assumes constant conditional variance. Goal counts are counts, and under the natural Poisson distribution model their variance equals their mean $x_i$, so observations with different expected goals are heteroscedastic. The reported homoscedastic standard error is therefore doubtful; a heteroscedasticity-robust calculation or a suitable count model would be more credible.

The fitted Poisson regression hasso that ${\mu }_i=e^{{\beta }_0}{x}_i^{{\beta }_1}$. Hypothesis $H_2$ requires both restrictionsas well as the adequacy of the assumed Poisson distribution. The displayed interval addresses only the single slope restriction ${\beta }_1=1$. Even if the separate interval for the intercept also contains zero, separate one-parameter intervals do not implement the relevant joint hypothesis test, because the two coefficient estimators can be correlated. One should test $({\beta }_0,{\beta }_1)=(0,1)$ jointly, for example by a likelihood-ratio test, and separately assess the Poisson goodness of fit. Thus the slope interval alone cannot decide whether $H_2$ is rejected at the $5\%$ level.

A Non-homogeneous Poisson point process with intensity function $\lambda$ is a random counting measure $N$ such that, for every finite family of pairwise disjoint bounded Borel sets $A_1,\dots ,A_k$, the counts $N(A_1),\dots ,N(A_k)$ are independent random variables andThe assumed local integrability makes each Poisson parameter finite. This count law and independence characterize the process.

Let $N$ be the homogeneous Poisson point process in ${\mathbb{R}}^3$ with intensity measure $\lambda dx$, and map each point $x$ to its radius $g(x)=|x|$. For an interval $(a,b]\subseteq {\mathbb{R}}_{+}$, the pushforward measure of the intensity isThe Mapping theorem for Poisson point processes therefore shows that the radii form a Non-homogeneous Poisson point process with intensity function

If $R$ is the distance to the closest star, the event $\{R>r\}$ says that the ball of radius $r$ contains no points. The zero-count probability of a Poisson distribution givesDifferentiating its cumulative distribution function yields the probability density function

Independence givesThis is the moment generating function of a Poisson distribution with mean $n$. Hence

Under a country-specific constant hazard survival model, let $D_k$ be the number of deaths and $Y_k$ the total person-time at risk in country $k$. The likelihood is proportional toso the maximum-likelihood estimator is ${\hat{\lambda }}_k=D_k/Y_k$. Test $H_0:{\lambda }_0={\lambda }_1$ by a likelihood-ratio test, score test, or Wald test; equivalently, give $D_k$ a Poisson distribution with mean ${\lambda }_k{Y}_k$ and test the country coefficient in a Poisson regression with offset $\log Y_k$.